Formatted question description: https://leetcode.ca/all/1116.html
1116. Print Zero Even Odd
Level
Medium
Description
Suppose you are given the following code:
class ZeroEvenOdd {
public ZeroEvenOdd(int n) { ... } // constructor
public void zero(printNumber) { ... } // only output 0's
public void even(printNumber) { ... } // only output even numbers
public void odd(printNumber) { ... } // only output odd numbers
}
The same instance of ZeroEvenOdd
will be passed to three different threads:
- Thread A will call
zero()
which should only output 0’s. - Thread B will call
even()
which should only ouput even numbers. - Thread C will call
odd()
which should only output odd numbers.
Each of the threads is given a printNumber
method to output an integer. Modify the given program to output the series 010203040506...
where the length of the series must be 2n.
Example 1:
Input: n = 2
Output: “0102”
Explanation: There are three threads being fired asynchronously. One of them calls zero(), the other calls even(), and the last one calls odd(). “0102” is the correct output.
Example 2:
Input: n = 5
Output: “0102030405”
Solution
This problem can be solved using semaphores. In the class, create three semaphores semaphoreZero
, semaphoreOdd
and semaphoreEven
for methods zero()
, odd()
and even()
respectively. Initially, semaphoreZero
has 1 permit, while semaphoreOdd
and semaphoreEven
has 0 permits.
For each method zero()
, odd()
and even()
, in the loop, there are three steps. The first step is to acquire a permit from the current semaphore. The second step is to call the printNumber.accept(x)
method for the current number. The third step is to release a permit back to the next semaphore.
-
import java.util.concurrent.Semaphore; import java.util.function.IntConsumer; public class Print_Zero_Even_Odd { // co-ordinate 3 threads public class ZeroEvenOdd { private int n; private Semaphore even; private Semaphore odd; private Semaphore zero; public ZeroEvenOdd(int n) { this.n = n; even = new Semaphore(0); odd = new Semaphore(0); zero = new Semaphore(1); } // printNumber.accept(x) outputs "x", where x is an integer. public void zero(IntConsumer printNumber) throws InterruptedException { for (int i = 0; i < n; i++) { zero.acquire(); printNumber.accept(0); if ((i & 1) == 0) { odd.release(); } else { even.release(); } } } public void even(IntConsumer printNumber) throws InterruptedException { for (int i = 2; i <= n; i += 2) { even.acquire(); printNumber.accept(i); zero.release(); } } public void odd(IntConsumer printNumber) throws InterruptedException { for (int i = 1; i <= n; i += 2) { odd.acquire(); printNumber.accept(i); zero.release(); } } } }
-
Todo
-
print("Todo!")